米勒-拉宾素数测试

由于卡米歇尔数的存在，导致 费马小定理 无法判断一个数是否是素数。

费马小定理: 设p是素数, a是任意整数且 a!三0( mod p )， 则

                     a^(p-1)  三  1(mod p)

 //========================================================

卡米歇尔数:它是合数, 当 1<=a<=n, 都有 a^n 三 a(mod n)

 //========================================================

卡米歇尔数的考塞特判别法: 设n是合数，则n是卡米歇尔数当且仅当它是奇数，且整除n的每个素数p满足下述两个条件:

1)p^2 不整除 n

2)p-1 整除 n-1

 //========================================================                  

如果要判断相当大的素数最好使用 合数的拉宾-米勒测试定理

 

合数的拉宾-米勒测试定理: 设n是奇素数, 记 n-1 = 2^k * q , q 是奇数, 对不被n整除的某个a, 如果下述两个条件都成立，则n是合数.

a) a^q !三 1（mod n);

b) 对所有 i = 0, 1, 2, ...., k-1,    a^((2^i)*q) !三 -1(mod n);

//========================================================  

 

这里给出了合数的拉宾-米勒测试定理  a  的取值:

 

if n < 1,373,653, it is enough to test a = 2 and 3.

if n < 9,080,191, it is enough to test a = 31 and 73.

if n < 4,759,123,141, it is enough to test a = 2, 7, and 61.

if n < 2,152,302,898,747, it is enough to test a = 2, 3, 5, 7, and 11.

// montgomery快速幂模算法 (n ^ p) % m, 与power算法极类似
unsigned __int64 montgomery(unsigned __int64 n, unsigned __int64 p, unsigned __int64 m)
{ 
    unsigned __int64 r = n % m;
    unsigned __int64 tmp = 1;
    while (p > 1)
    {
        if ((p & 1)!=0)
        {
            tmp = (tmp * r) % m;
        }
        r = (r * r) % m;
        p >>= 1;
    }
    return (r * tmp) % m;
}


//返回true:n是合数, 返回false:n是素数
bool R_M_Help(unsigned __int64 a, unsigned __int64 k, unsigned __int64 q, unsigned __int64 n)
{
	if ( 1 != montgomery( a, q, n ) )
    {
        int e = 1;
        for ( int i = 0; i < k; ++i )
        {
            if ( n - 1 == montgomery( a, q * e, n ) ) 
                return false;
           
            e <<= 1;
        }
        
		return true;
    }


	return false;
}


//拉宾-米勒测试 返回true:n是合数, 返回false:n是素数  
bool R_M( unsigned __int64 n ) 
{
	if( n < 2 )
		throw 0;


	if ( n == 2 || n == 3 )
	{
		return false;
	}


	if( (n & 1) == 0 )
		return true;


    // 找到k和q, n = 2^k * q + 1;
    unsigned __int64 k = 0, q = n - 1;
    while( 0 == ( q & 1 ) )
    {
        q >>= 1;
        k++;
    }


	/*if n < 1,373,653, it is enough to test a = 2 and 3. 
	if n < 9,080,191, it is enough to test a = 31 and 73. 
	if n < 4,759,123,141, it is enough to test a = 2, 7, and 61. 
	if n < 2,152,302,898,747, it is enough to test a = 2, 3, 5, 7, and 11.*/
         
	if( n < 1373653 )
	{
		if( R_M_Help(2, k, q, n ) 
		 || R_M_Help(3, k, q, n ) )
			return true;
	}
	else if( n < 9080191 )
	{
		if( R_M_Help(31, k, q, n ) 
		 || R_M_Help(73, k, q, n ) )
			return true;
	}  
	else if( n < 4759123141 )
	{
		if( R_M_Help(2, k, q, n ) 
		 || R_M_Help(3, k, q, n )
		 || R_M_Help(5, k, q, n )
		 || R_M_Help(11, k, q, n ) )
			return true;
	}
	else if( n < 2152302898747 )
	{
		if( R_M_Help(2, k, q, n ) 
		 || R_M_Help(3, k, q, n )
		 || R_M_Help(5, k, q, n )
		 || R_M_Help(7, k, q, n )
		 || R_M_Help(11, k, q, n ) )
			return true;
	}
	else 
	{
		if( R_M_Help(2, k, q, n ) 
		 || R_M_Help(3, k, q, n )
		 || R_M_Help(5, k, q, n )
		 || R_M_Help(7, k, q, n )
		 || R_M_Help(11, k, q, n )
		 || R_M_Help(31, k, q, n )
		 || R_M_Help(61, k, q, n )
		 || R_M_Help(73, k, q, n ) )
			return true;
	}


    return false;
}

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<time.h>
#include<stdlib.h>
long long prime[]={2,3,5,7,11,13,17,19,23,29};
#define TIME 10    //Miller测试次数
long long gcd(long long a,long long b){return b==0?a:gcd(b,a%b);}
long long mod_mult(long long a, long long b, long long n) //计算(a*b) mod n
{
    long long s=0; a=a%n;
    while(b){
        if (b&1){
            s += a;if(s>=n) s-= n;
        }
        a=a<<1;if(a>=n)a-=n; b=b>>1;
    }
    return s;
}
long long mod_exp(long long a, long long b, long long n) //计算(a^b) mod n
{
    long long d=1; a=a%n;
    while(b>=1) {
        if(b&1)d=mod_mult(d,a,n);
        a=mod_mult(a,a,n); b=b>>1;
    }
    return d;
}

bool Wintess(long long a, long long n) //以a为基对n进行Miller测试并实现二次探测
{
    long long m, x, y;
    int i, j = 0; m = n-1;
    while (!(m&1)) //计算(n-1)=m*(2^j)中的j和m,j=0时m=n-1,不断的除以2直至m为奇数
    {
        m=m>>1; j++;
    }
    x = mod_exp(a, m, n);
    for (i = 1; i <= j; i++) {
        y = mod_mult(x, x, n);
        if ((y == 1) && (x != 1) && (x != n - 1)) //二次探测
            return true; //返回true时,n是合数
        x = y;
    }
    if (y!=1) return true;
    return false;
}
bool miller_rabin(int times, long long n) //对n进行times次的Miller测试
{
    long long a;
    int i;
    if (n == 2) return true;
    if (n<2||(n&1)==0) return false;
    for (i = 1; i <= times; i++) {
        a =rand() % (n - 2) + 2;
        if (Wintess(a, n)) return false;
    }
    return true;
}